Expert: Paul Klarreich - 11/20/2010

Question
QUESTION: Dear Mr.Paul,
      In the image which I have uploaded(has the question.) And my answers to this question are as follows:
y = 12cos(1.5x)-3  and
y = 12sin{1.5(x-pi)}-3.
I just want to make sure whether they are correct or not and how can i make them better according to your opinion.
         Thanks a lot,
         Bulbul.

ANSWER: Questioner:   Bulbul
Country:  Canada
Category:  Advanced Math
Private:  No
 
Subject:  Trigonometry(Graphs)
Question:  

In the image which I have uploaded(has the question.) And my
answers to this question are as follows:
y = 12cos(1.5x)-3  and
y = 12sin{1.5(x-pi)}-3.
I just want to make sure whether they are correct or not and how can

i make them better according to your opinion.
         Thanks a lot,
         Bulbul.  
---------------------------------------------------------
here is the process:

0. the amplitude is one-half of the difference:  (12 - (-15))/2

= 27/2 = 13.5

1. The period appears to be P = 4.  That means Bx = 2pi when x = 4.
Solve that to get B = pi/2

2. The vertical offset is the average of +12 and -15.  
  That is not -3, is it?

3C:  For the cosine version, you have your max at x = 0, so there is
no phase shift. (it is zero)

3S: For the sine version, you have your zero at -1, so solve:

B(x - phi) = 0

for phi.  Do this:

pi/2(-1 - phi) = 0

phi = -1.

Put that together and I think you should have:

y = 13.5 cos(pi/2 (x)) - 1.5

y = 13.5 sin(pi/2(x + 1)) - 1.5

In my opinion, fixing the amplitude, displacement, phase, and
frequency makes them are better.  (Well, you asked for my opinion,
didn't you?)


---------- FOLLOW-UP ----------

QUESTION: Dear Mr.Paul,
       I am sorry to disturb you again but the amplitude and vertical displacement which you told me to change should not remain same as I did? Because the maximum which you took ie 12 is indeed 9 in te graph;then the amplitude would change to 12 and vertical displacement to -3. Am I not correct,Mr.Paul according to the graph? I have also uploaded the image again and have marked the maximum.So,what do you say,Mr.Paul?
         Thanks a lot,
         Bulbul.

Answer
QUESTION: Dear Mr.Paul,
      I am sorry to disturb you again but the amplitude and vertical displacement which you told me to change should not remain same as I did?

>> No. The amplitude is 13.5, not 12. The V.D. is -1.5, not 3.

Because the maximum which you took ie 12 is indeed 9 in te graph;then the amplitude would change to 12 and vertical displacement to -3. Am I not correct,Mr.Paul according to the graph? I have also uploaded the image again and have marked the maximum.So,what do you say,Mr.Paul?

>> I say I saw it the first time.  

Think about this: Suppose the graph were displaced another 100 (yes, 100) units down.  Would the amplitude be negative 88?

The amplitude is the distance from the center of the oscillation to either its max or min.  You cannot have a 'positive' amplitude of 12 and a 'negative' amplitude of 15.  You have ONE amplitude.

Here is a hint:  Find some graphing program on the web, feed in the function, see what you get.



         Thanks a lot,
         Bulbul.